This set of Probability and Statistics Assessment Questions and Answers focuses on “Set Theory of Probability – 2”.
1. If P(B⁄A) = p(b), then P(A ∩ B) = ____________
a) p(b)
b) p(a)
c) p(b).p(a)
d) p(a) + p(b)
View Answer
Explanation: P(B/A) = p(b) implies A and B are independent events
Therefore, P(A ∩ B) = p(a).p(b).
2. Two unbiased coins are tossed. What is the probability of getting at most one head?
a) 1⁄2
b) 1⁄3
c) 1⁄6
d) 3⁄4
View Answer
Explanation: Total outcomes = (HH, HT, TH, TT)
Favorable outcomes = (TT, HT, TH)
At most one head refers to maximum one head,
Therefore, probability = 3⁄4.
3. If A and B are two events such that p(a) > 0 and p(b) is not a sure event, then P(A/B) = ?
a) 1 – P(A /B)
b) \(P\frac{(\bar{A})}{(\bar{B})}\)
c) Not Defined
d) \(\frac{1-P(A \cup B)}{P(\bar{B}}\)
View Answer
Explanation: From definition of conditional probability we have
\(P(\bar{A}/\bar{B})=\frac{\bar{A}\cap\bar{B}}{P(\bar{B})}\)
Using De Morgan’s Law
=\(\frac{P(\bar{A \cup B})}{(\bar{B}}\)
=\(\frac{1-P(A \cup B)}{P(\bar{B}}\)
4. If A and B are two events, then the probability of exactly one of them occurs is given by ____________
a) P(A ∩ B) + P(A ∩ B)
b) P(A) + P(B) – 2P(A) P(B)
c) P(A) + P(B) – 2P(A) P(B)
d) P(A) + P(B) – P(A ∩ B)
View Answer
Explanation: The set corresponding to the required outcome is
(A ∩ B) ∪ (A ∩ B)
Hence the required probability is
P(P(A ∩ B) ∪ (A ∩ B)) = (A ∩ B) + P(A ∩ B).
5. The probability that at least one of the events M and N occur is 0.6. If M and N have probability of occurring together as 0.2, then P(M) + P(N) is?
a) 0.4
b) 1.2
c) 0.8
d) Indeterminate
View Answer
Explanation: Given : P(M ∪ N) = 0.6, P(M ∩ N) = 0.2
P(M ∪ N) + P(M ∩ N) = P(M) + P(N)
2 – (P(M ∪ N) + P(M ∩ N)) = 2 – (P(M) + P(N))
= (1 – P(M)) + (1 – P(N))
2 – (0.6 + 0.2) = P(M) + P(N)
P(M) + P(N) = 2 – 0.8
= 1.2
6. A jar contains ‘y’ blue colored balls and ‘x’ red colored balls. Two balls are pulled from the jar without replacing. What is the probability that the first ball Is blue and second one is red?
a) \(\frac{xy-y}{x^2+y^2+2xy-(x+y)}\)
b) \(\frac{xy}{x^2+y^2+2xy-(x+y)}\)
c) \(\frac{y^2-y}{x^2+y^2+2xy-(x+y)}\)
d) \(\frac{xy-y}{x^2+y^2+2xy-(x-y)}\)
View Answer
Explanation: Number of blue balls = y
Number of Red balls = x
Total number of balls = x + y
Probability of Blue ball first = y / x + y
No. of balls remaining after removing one = x + y – 1
Probability of pulling secondball as Red=\(\frac{x}{x+y-1}\)
Required porbability=\(\frac{y}{(x+y)}\frac{x}{(x+y-1)}=\frac{xy}{x^2+y^2+2xy-(x+y)}\)
7. A survey determines that in a locality, 33% go to work by Bike, 42% go by Car, and 12% use both. The probability that a random person selected uses neither of them is?
a) 0.29
b) 0.37
c) 0.61
d) 0.75
View Answer
Explanation: Given: p(b) = 0.33, P(c) = 0.42
P(B ∩ C) = 0.12
P(B ∩ C) = ?
P(B ∩ C) = 1 – P(B ∪ C)
= 1 – p(b) – P(c) + P(B ∩ C)
= 1 – 0.22 – 0.42 + 0.12
= 0.37.
8. A coin is biased so that its chances of landing Head is 2⁄3. If the coin is flipped 3 times, the probability that the first 2 flips are heads and the 3rd flip is a tail is?
a) 4⁄27
b) 8⁄27
c) 4⁄9
d) 2⁄9
View Answer
Explanation: Required probability = 2⁄3 x 2⁄3 x 1⁄3 = 4⁄27.
9. Husband and wife apply for two vacant spots in a company. If the probability of wife getting selected and husband getting selected are 3/7 and 2/3 respectively, what is the probability that neither of them will be selected?
a) 2⁄7
b) 5⁄7
c) 4⁄21
d) 17⁄21
View Answer
Explanation:Let H be the event of husband getting selected
W be the event of wife getting selected
Then, the event of neither of them getting selected is = (H ∩ W)
P (H ∩ W) = P (H) x P (W)
= (1 – P (H)) x (1 – P (W))
= (1 – 2⁄3) x (1 – 3⁄7)
= 4⁄21.
10. For two events A and B, if P (B) = 0.5 and P (A ∪ B) = 0.5, then P (A|B) = ?
a) 0.5
b) 0
c) 0.25
d) 1
View Answer
Explanation: We know that,
P (A│B) = P(A ∩ B)/P(B)
= P((A ∪ B)/P(B))
= (1 – P(A ∪ B)) /P(B)
= (1 – 0.5)/0.5
= 1.
11. A fair coin is tossed thrice, what is the probability of getting all 3 same outcomes?
a) 3⁄4
b) 1⁄4
c) 1⁄2
d) 1⁄6
View Answer
Explanation:S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
All 3 same outcomes mean either all head or all tail
Total outcomes = 8
Favourable outcomes = {HHH, TTT} = 2
∴ required probability = 2⁄8 = 1⁄4.
12. A bag contains 5 red and 3 yellow balls. Two balls are picked at random. What is the probability that both are of the same colour?
a) \(\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}\)
b) \(\frac{C_2^5 * C_2^3}{C_2^8}\)
c) \(\frac{C_1^5 * C_1^3}{C_2^8}\)
d) 0.5
View Answer
Explanation:Total no.of balls = 5R+3Y = 8
No.of ways in which 2 balls can be picked = 8C2
Probability of picking both balls as red = 5C2 /8C2
Probability of picking both balls as yellow = 3C2 /8C2
∴ required probability \(\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}\).
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